Timeline for answer to Is the least common multiple sequence $\text{lcm}(1, 2, \dots, n)$ a subset of the highly abundant numbers? by GH from MO
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| Dec 17, 2025 at 18:05 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 7, 2025 at 0:45 | comment | added | GH from MO | @SaúlRM Correcting a sentence in my previous comment: If you look at the last inequality in the proof of the Theorem in my post, you will see that it gets weaker (hence easier to satisfy) as $s_j$ increases. | |
| Oct 6, 2025 at 22:04 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 6, 2025 at 21:52 | comment | added | Saúl RM | @GHfromMO oh, $\sum\frac{1}{p_j}>2$ is the sum that I had found wrongly yesterday. It seems that estimate coincidentally turned out to be right, except that instead of $p-1$ not being multiple of the squares, it has to be multiple of the eight powers | |
| Oct 6, 2025 at 21:43 | comment | added | GH from MO | @SaúlRM (continued) it is sufficient to require that $p-1$ is divisible by the eighth power of the first $59$ primes. A good compromise (or trade-off) is that $p-1$ is divisible by the fourth power of the first $70$ primes. The solution of this optimization problem is that the $s_j$'s are not equal. But again, this is not so interesting now, because Terry's construction is more effective. | |
| Oct 6, 2025 at 21:40 | comment | added | GH from MO | @SaúlRM If you look at the last inequality in the proof of the Theorem in my post, you will see that it gets stronger as $s_j$ increases. If the $s_j$'s tend to infinity, then the condition says that $\sum_j 1/p_j$ exceeds $2$. This latter (weaker) condition is satisfied if the $p_j$'s include the first $59$ primes. In fact if each $s_j$ is at least $8$, and the $p_j$'s include the first $59$ primes, then the sum (in the last display of the proof of the Theorem) already exceeds $2$. Hence instead of requiring that $p-1$ is divisible by the first thousand primes (actually $960$ primes suffice), | |
| Oct 6, 2025 at 17:35 | comment | added | Saúl RM | @GHfromMO I think you are right; as I mentioned in a now deleted comment, I was thinking an inequality went in the opposite direction, it is actually better if the number is very non-squarefree | |
| Oct 6, 2025 at 16:38 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 6, 2025 at 15:14 | comment | added | GH from MO | @SaúlRM I realized that higher prime powers in $p−1$ do help a little bit, see the Remark after the Theorem. This allows to reduce the astronomical bounds considerably, but they still remain astronomical of course. Terry's approach in his second response is the way to go. | |
| Oct 6, 2025 at 14:49 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 6, 2025 at 14:35 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 6, 2025 at 4:47 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 20:50 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 20:26 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 20:16 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 19:13 | comment | added | Matthew Bolan | I definitely did observe a rather dramatic reduction when "cutting out the middle man" to create the new holdouts list. For example, for $n \in (1108801, 1771561)$ and $p$ the largest prime $\le n$ with $p \equiv 1 \pmod {2310}$, I found that $\sigma(L_n (p-1)/p) \ge \sigma(L_n)$. | |
| Oct 5, 2025 at 18:58 | comment | added | Matthew Bolan | Well, there's little reason to create a new prime ladder right now, but my code is fast enough that implementing some amount of @SaúlRM 's suggestion might make it worth it. | |
| Oct 5, 2025 at 18:38 | comment | added | GH from MO | @MatthewBolan I updated the Theorem. The original proof contained a silly calculational error, which is fixed now. Unfortunately, the new proof requires that $p-1$ is divisible by much more primes, e.g. the first thousand primes. I am sorry about this. | |
| Oct 5, 2025 at 18:35 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 18:19 | comment | added | GH from MO | @SaúlRM Actually, this error will cost us a lot, because there is a big difference between $\sum_j 1/(p_j-1)$ and $\sum_j (p_j-1)/p_j^2$. Both can be pushed over $2$, but the latter sum needs much more terms for this to happen. Please say thanks to Golvano for me. | |
| Oct 5, 2025 at 18:08 | comment | added | Saúl RM | @GHfromMO no problem. I found this after Eduardo Rodríguez Golvano told me something looked amiss in the proof | |
| Oct 5, 2025 at 18:05 | comment | added | GH from MO | @SaúlRM You are absolutely right. See my previous comment - and thank you! | |
| Oct 5, 2025 at 18:05 | comment | added | GH from MO | @MatthewBolan Yes, there is a small calculational error in my proof, I got confused with division in the denominator. Thanks to you and Saúl RM for spotting this. I will update my proof shortly. The statement will change a bit, unfortunately. | |
| Oct 5, 2025 at 17:51 | comment | added | Saúl RM | @GHfromMO one thing, when dividing both sides by $p-1$, in the previous expression the $(1−p_j^{−1})$ is multiplying/in the numerator, but in the next step the $(p_j-1)$ appears in the denominator. Is there a problem there? Even if I am right this would not kill the proof, just make the bounds worse (I think). Also, nice argument! I just got around to reading it in detail. For some reason I had thought that $p-1$ being squarefree was necessary, but indeed it is completely unnecessary | |
| Oct 5, 2025 at 17:48 | comment | added | Matthew Bolan | When writing the code to list holdouts I encountered the example of $n=p=32341$, which seems to meet the conditions of your theorem, and yet I find $\sigma(L_{n} \cdot (p-1) / p) < \sigma(L_n)$. | |
| Oct 5, 2025 at 17:36 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 16:19 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 16:13 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 0:14 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 5, 2025 at 0:06 | comment | added | Matthew Bolan | In that case, I get that $p$ handles all $n$ in $[p, \frac{121}{120} (p-1)]$, and then checking every $p \equiv 1 \pmod {2310}$ I find that the largest prime power less than $10^{10}$ that cannot be handled is $3229379$. | |
| Oct 5, 2025 at 0:02 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 4, 2025 at 23:54 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 4, 2025 at 23:47 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 4, 2025 at 23:38 | comment | added | Matthew Bolan | A quick computation suggests that this handles every $n$ greater than 3202690 and less than $10^{10}$ | |
| Oct 4, 2025 at 23:25 | comment | added | Terry Tao | Nice, so every prime $p = 1 \hbox{ mod } 2310$ can treat the interval $n \in [\frac{120}{121}(p-1), p]$. That should be able to rapidly cover all medium-sized $n$, and some effective prime number theorem in arithmetic progressions, eg arxiv.org/abs/2301.13457 assuming GRH, could handle larger $n$. Without GRH we have arxiv.org/abs/1305.3087 but it may take some (standard) effort to convert this to a usable PNT in short intervals. | |
| Oct 4, 2025 at 23:12 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 4, 2025 at 19:59 | comment | added | Matthew Bolan | Ok @TerryTao , I wrote up something short and linked my code at mathoverflow.net/a/501192/177778. It turns out that if one uses 1.004 instead of 1.0045 the search space is as small as 589824 even without any pruning. As a sanity check the code is also able to prove $L_{71}$ is not highly abundant, and finds the exact example already in this thread (and no others). | |
| Oct 4, 2025 at 19:29 | comment | added | José Damián Espinosa | @TerryTao and GH from MO, my real name is Damian, a real apology for the name "283212123", I've been wanting to change it for a few days, but Mathoverflow only lets me exactly in for 4 days, in 4 days when Mathoverflow finally allows me, I will change it... | |
| Oct 4, 2025 at 18:27 | comment | added | José Damián Espinosa | @TerryTao, I have ventured to verify all the combinations that Bolan propose, at the moment I have 238910000, and none $\sigma(m)$ exceeds $\sigma(L_{67})$, probably it took me about 17 more hours verify all the 1509949440 combinations! | |
| Oct 4, 2025 at 16:43 | comment | added | Terry Tao | Alternatively, once the exponent ranges are determined, the problem becomes (after taking logarithms) a 26-dimensional integer program, with many variables tightly constrained, and I would imagine that standard integer programming packages could also work here (although one then has to trust those packages to be bug-free). | |
| Oct 4, 2025 at 16:36 | comment | added | Matthew Bolan | @TerryTao yes, I just filtered out products larger than $L_{67}$ and this was enough. I will post details later today. | |
| Oct 4, 2025 at 16:32 | comment | added | Terry Tao | I'd also be interested in seeing the details. A back of the envelope calculation indicates to me that @MatthewBolan's exponent ranges cut the number of cases down to 1509949440, but this is still somewhat large for a computer verification, though not impossible. Did you take further advantage of the score function (or filtering out those products larger than $L_{67}$) to cut the number of cases down further? | |
| Oct 4, 2025 at 15:59 | comment | added | GH from MO | @283212123 If you believe the comments of Tao and Bolan above, there are better ways to check that $L_{67}$ is highly abundant than to go through all $m$'s up to $L_{67}$ and calculate their sums of divisors. I asked Bolan to provide full details of his claim that $L_{67}$ is highly abundant. | |
| Oct 4, 2025 at 15:51 | comment | added | José Damián Espinosa | @GHfromMO, It is really difficult, the only way is to verify that $\sigma(L_{67})>\sigma(m)$ for absolutely all $m<n$, and it is absolutely impossible!!, there are no more useful criteria right, than the definition itself? | |
| Oct 4, 2025 at 12:23 | comment | added | GH from MO | @MatthewBolan Can you add a detailed proof as a response (not as a comment) here? Together with existing responses, this would show that $L_{71}$ is the smallest counterexample to the OP's conjecture (which is kind of cool). | |
| Oct 4, 2025 at 7:03 | comment | added | Matthew Bolan |
Using the score function of @TerryTao I find that no primes greater than $109$ can occur in the prime factorization of $M$, and get the following inclusive ranges for the exponents {2: (5, 12), 3: (3, 7), 5: (2, 4), 7: (1, 3), 11: (1, 2), 13: (1, 2), 17: (1, 2), 19: (1, 2), 23: (1, 1), 29: (1, 1), 31: (1, 1), 37: (0, 1), 41: (0, 1), 43: (0, 1), 47: (0, 1), 53: (0, 1), 59: (0, 1), 61: (0, 1), 67: (0, 1), 71: (0, 1), 73: (0, 1), 79: (0, 1), 83: (0, 1), 89: (0, 1), 97: (0, 1), 101: (0, 1), 103: (0, 1), 107: (0, 1), 109: (0, 1)}. Checking these, I find that $L_{67}$ is highly abundant.
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| Oct 4, 2025 at 3:33 | comment | added | Terry Tao | I'm beginning to suspect that $L_{67}$ is in fact highly abundant. Define the "score" of $M$ to be $\sigma(M)/M^{1.0045}$. If there is to be an $M$ for which $M < L_{67}$ and $\sigma(M) \geq \sigma(L_{67})$, then $M$ has to have a higher score than $L_{67}$. But of all the ways to add and remove primes from $L_{67}$, only the addition of $2$, $3$, or $3^2$, or the deletions of $59$, $61$, or $67$, increase the score; all other moves decrease the score, often by a lot (i.e. $L_{67}$ is nearly collossally abundant). This should reduce things to a feasible case check. | |
| Oct 4, 2025 at 3:12 | comment | added | Alexis Olson | Best near miss I can find: $L := L_{67} = 2^6 \cdot 3^3 \cdot 5^2 \cdot 11 \cdot 59 \cdot 67 \cdot L',$ and $M := 2^9 \cdot 3^5 \cdot 5^3 \cdot 11^2 \cdot L'.$ Off by only $0.177\%$. | |
| Oct 4, 2025 at 1:11 | comment | added | Terry Tao | Interesting that the newest example ($67 \times 71 = 4757$ versus $2^2 \times 3 \times 5 \times 79 = 4740$) uses a prime larger than $n$ for the first time. So it's all down to $L_{67}$ now... | |
| Oct 4, 2025 at 1:06 | comment | added | GH from MO | @TerryTao The latest example is $n=71$. (Sorry for making the earlier examples disappear.) | |
| Oct 4, 2025 at 1:05 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 4, 2025 at 1:00 | comment | added | Terry Tao | [This was for the $n=97$ example in the edit history; for $n=79$, the numbers are instead $67 \times 79 = 5293$ and $2^5 \times 3 \times 5 \times 11 = 5280$. For $n=113$, it was $89 \times 113 = 10051$ versus $2 \times 5 \times 7 \times 11 \times 13 = 10010$.] | |
| Oct 4, 2025 at 0:58 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 4, 2025 at 0:45 | comment | added | Terry Tao | The key here in the second response seems to be that the rough number $89 \times 97 = 8633$ (which is the product of primes only slightly less than or equal to $n$) is slightly larger than the smooth number $2^2 \times 3 \times 5 \times 11 \times 13 = 8580$ (which is the product of primes slightly larger than a root $n^{1/j}$ of $n$). There doesn't seem to be much more "room" for any smaller example of this type, though. | |
| Oct 3, 2025 at 22:14 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 3, 2025 at 21:53 | history | edited | LSpice | CC BY-SA 4.0 |
Link to counterexample
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| Oct 3, 2025 at 21:37 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 3, 2025 at 9:30 | comment | added | GH from MO | @DanielWeber So my algorithm is not deterministic in the sense that it starts from a random tuple $\mathcal{Q}$, and it tries to modifiy it into a locally minimizing tuple $\mathcal{Q}$. By running the algorithm several times, and tweaking the parameters, better and better pairs of tuples $(\mathcal{P},\mathcal{Q})$ arose, which I kept recording manually in a text file. Hope this helps! | |
| Oct 3, 2025 at 9:16 | comment | added | GH from MO | @DanielWeber My code is not professionally written, so I would be ashamed to share it, but I can tell you how it works. It starts with a given consecutive 15-tuple of primes $\mathcal{P}$, and it picks a random 8-tuple of primes $\mathcal{Q}$ from a given interval such that $\prod\mathcal{Q}>\prod\mathcal{P}$. It checks if $\mathcal{Q}$ can be updated by exchanging one if its elements, to decrease $\prod\mathcal{Q}$ but still maintaining $\prod\mathcal{Q}>\prod\mathcal{P}$. This last step is repeated several hundred times. For the final $\mathcal{Q}$, Terry's $(1)$ and $(2)$ are displayed. | |
| Oct 3, 2025 at 6:28 | comment | added | Daniel Weber | Would you mind sharing your code? | |
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| Oct 2, 2025 at 23:07 | history | answered | GH from MO | CC BY-SA 4.0 |